Suppose $a_0,a_1,b,c,d$ with $a_0 \neq a_1$ are integers, or elements of any ${\mathbb Z}_p$ for prime $p$.
A while ago I was attempting to find a set of such numbers that satisfied the equations
$$a_0 a_0 = ba_0 + ca_0 + d$$ $$a_0 a_1 = ba_0 + ca_1 + d$$ $$a_1 a_1 = ba_1 + ca_1 + d$$
and after many failed attempts ultimately conjectured that it was impossible.
I set about trying to prove this, but was immediately swamped in equations and ring axioms, none of which availed any insights.
I also noted that the equations can be satisfied in integer rings. For example, $a_0 = 2, a_1 = 5, b,c = 8, d = 8$ in ${\mathbb Z}_9$.
Can anyone provide insights as to how to prove the conjecture? Or is it in fact wrong, and can somebody procure a set of numbers that satisfies the equations in the specified fields?
Many thanks for any assistance.
Edit: I should note that the original problem I looked at had the additional condition that $$a_1 a_0 = ba_1 + ca_0 + d$$ which trivially requires that $b = c$.
This case is also handled by the supplied answer, however.
Assume we are in an integral domain.
If we subtract the first and third equations, we get $a_0^2 - a_1^2 = b a_0 + c a_0 - b a_1 - c a_1$, which is the same as $(a_0 - a_1)(a_0 + a_1) = (b + c)(a_0 - a_1)$. Since $a_0 \ne a_1$, $a_0 - a_1 \ne 0$, and so we can cancel to get $a_0 + a_1 = b + c$.
Subtracting the second and third, we get $a_0 a_0 - a_0 a_1 = c a_0 - c a_1$, which, after cancellation, becomes $a_0 = c$. This tells us that $a_1 = b$.
Now our equations have fewer variables, and we can see the redundancy: $$ c^2 = bc + c^2 + d $$ $$ bc = bc + cb + d $$ $$ b^2 = b^2 + cb + d $$
All three tell us that $d = -bc$. The solution set can be parameterized as $(a_0, a_1, b, c, d) = (t, u, u, t, -tu)$. As an example, $(1, 2, 2, 1, -2)$ works:
$$ 1^2 = 2 \cdot 1 + 1 \cdot 1 + (-2) $$ $$ 1 \cdot 2 = 2 \cdot 1 + 1 \cdot 2 + (-2) $$ $$ 2^2 = 2 \cdot 2 + 1 \cdot 2 + (-2) $$